Find the largest possible integer pair (m, n) which satisfy (m2-mn –n2)2=1. where m, n?

Answer 1

There is no largest pair that satisfies this.
We will prove this, and, in fact, show
that all the n which satisfy this equation are
the Fibonacci numbers.
The given equation implies that
1). m² -mn-n² = -1
or
2). m²-mn-n² = 1.
I'll give the details for case 1 and just repeat
the essentials for case 2.
m² -mn-(n²-1) = 0
This is a quadratic equation and, since
we want it to have integer solutions, its
discriminant, n² + 4(n²-1) must be a square, call it x².
So x² - 5n² = -4. (*)
This is a Pell equation, so there are indeed
infinitely many solutions to it. Let's find them.
It turns out that all the integer solutions to this equation
are given by
(x, n) = [(1+√5)/2]^k,
where k is odd.
But we can prove by induction that
[(1+√5)/2]^k = (L_k + F_k√5)/2
for ALL k, where F_k is the kth Fibonacci number
and L_k is the kth Lucas number, defined by
L_1 = 1, L_2 = 3, L_(n+1) = L_n + L_(n-1).
So all the solutions of (*) are
given by (x,n) = (L_k, F_k), where k is odd.
Sorry to leave out the gory details of all
the proofs, but typing math in this setting
is just awful!
For the second case we get
x² - 5n² = 4,
and all the solutions to this equation are given by
(x,n) = (L-k, F_k), where k is even.
So the set of n satisfying the original equation
is the set of Fibonacci numbers.